tìm các số x , y , z biết rằng : 3x = 4y , 5y = 6z va xyz = 30
tìm x : \(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
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ta có :3x=4y,5y=6z
=>\(\dfrac{x}{4}\)=\(\dfrac{y}{3}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)=k
=> x=8k ; y=6k ; z=5k
=> 8k.6k.5k=30
=> 240k3 =30
=>k3 =8
=>k=2
=> x=8.2=16 ; y=6.2=12 ; x =5.2=10
Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
Ta có:
\(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}=xyz\left(\frac{1}{yz\left(1+x^2\right)}+\frac{18}{xz\left(1+y^2\right)}+\frac{4}{xy\left(1+z^2\right)}\right)\)
\(=xyz\left(\frac{1}{yz+x\left(x+y+z\right)}+\frac{18}{xz+y\left(x+y+z\right)}+\frac{4}{xy+z\left(x+y+z\right)}\right)\)
\(=xyz\left(\frac{1}{\left(x+y\right).\left(x+z\right)}+\frac{18}{\left(y+x\right).\left(y+z\right)}+\frac{4}{\left(z+x\right).\left(z+y\right)}\right)\)
\(=xyz.\frac{\left(z+y\right)+18.\left(x+z\right)+4\left(x+y\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)
\(=\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)(đpcm)
Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=k\Rightarrow x=3k;y=5k;z=7k\)
\(xy+yz+zx=3k.5k+5k.7k+7k.3k=k^2\left(15+35+21\right)=71k^2;xyz=3k.5k.7k=105k^3\)
Ta có : \(xyz\left(xz+yz+xy+xz+yz+xy\right)=477120\)
\(\Rightarrow xyz\left(xz+yz+xy\right)=238560\)\(\Rightarrow105k^3.71k^2=238560\Rightarrow k^5=32=2^5\Rightarrow k=2\)
Vậy : x= 6 ; y = 10 ; z = 14
\(\Sigma\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\Sigma\left(\dfrac{1}{9}.\dfrac{a^2\left(2+1\right)^2}{2a.\left(\Sigma a\right)+2a^2+bc}\right)\le\Sigma\left(\dfrac{1}{9}.\dfrac{4a^2}{2a\left(\Sigma a\right)}+\dfrac{1}{9}.\dfrac{a^2}{2a^2+bc}\right)\)
\(=\Sigma\left(\dfrac{1}{9}.\left(\dfrac{2a}{\Sigma a}+\dfrac{a^2}{2a^2+bc}\right)\right)=\dfrac{1}{9}\left(2+\Sigma\dfrac{a^2}{2a^2+bc}\right)\)
Cần chứng minh \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
<=> \(\Sigma\frac{bc}{2a^2+bc}\ge1\) (*)
Đặt (x;y;z) -------> \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Suy ra (*) <=> \(\Sigma\frac{x^2}{x^2+2xy}\ge1\Leftrightarrow\frac{\Sigma x^2}{\Sigma x^2}\ge1\) (đúng)
Vậy \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
Suy ra \(\Sigma\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\le\frac{1}{9}\left(2+\Sigma\frac{a^2}{2a^2+bc}\right)\le\frac{1}{9}\left(2+1\right)=\frac{1}{3}\)
Đẳng thức xảy ra <=> x = y = z = 1
\(ĐK:x,y,z\ne0\)
Đặt \(6\left(x-\frac{1}{y}\right)=3\left(y-\frac{1}{z}\right)=2\left(z-\frac{1}{x}\right)=xyz-\frac{1}{xyz}=a\)
\(\Rightarrow x-\frac{1}{y}=\frac{a}{6};y-\frac{1}{z}=\frac{a}{3};z-\frac{1}{x}=\frac{a}{2}\)\(\Rightarrow\frac{a^3}{36}=xyz-\frac{1}{xyz}-x+\frac{1}{y}-y+\frac{1}{z}-z+\frac{1}{x}=a-\frac{a}{6}-\frac{a}{3}-\frac{a}{2}=0\)suy ra a = 0
Nếu xyz = 1 thì x = y = z = 1 (thỏa mãn)
Nếu xyz = -1 thì x = y = z = -1 (thỏa mãn)
Vậy nghiệm của hệ phương trình (x; y; z) là: (1; 1; 1),(-1; -1; -1).
1)
a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )
5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)
\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)
b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)
\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)
\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy ...
\(1)\) Ta có :
\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)
\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xyz=30\) ta được :
\(8k.6k.5k=30\)
\(\Leftrightarrow\)\(240k^3=30\)
\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)
\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)
\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\)\(k=\frac{1}{2}\)
Suy ra :
\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)
\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)
\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)
Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)
Chúc bạn học tốt ~